Degree of a Vertex (Advanced Applications and Extensions)

False, because in a simple digraph a vertex cannot have an edge to itself, so the maximum possible out-degree is n − 1 rather than n.

An isolated vertex has no incident edges in either direction, giving both in-degree and out-degree of 0. A sink has in-degree greater than 0 but out-degree 0. A source has out-degree greater than 0 but in-degree 0. A leaf has degree 1 in undirected graphs and refers to a tree endpoint.

In a simple graph degree counts distinct neighbors since no loops or multiple edges exist. Degree 5 means exactly 5 neighboring vertices. With 6 vertices total the only vertex A cannot connect to is itself, confirming 5 neighbors. Option D would require a loop which is forbidden in simple graphs.

Each vertex connects to at most n-1 others giving n(n-1) ordered connections. Since undirected edges are unordered each is counted twice in that total, so dividing by 2 gives n(n-1)/2 unique edges. Option A ignores the n-1 factor. Option B uses n+1 instead of n-1. Option D fails to divide by 2.

Degree 2 at every vertex means each vertex lies on exactly two edges with no branching and no endpoints. This forces edges to connect in closed loops. Trees have leaves of degree 1. Paths have two endpoints of degree 1. Stars have one center of high degree. Only cycles give every vertex exactly degree 2.

A source emits edges but receives none, functioning as a starting point in flow or dependency graphs. A sink is the opposite — it receives edges but emits none. An isolated vertex has both in-degree and out-degree equal to 0. A universal vertex is adjacent to all others, which requires positive in-degree.

Each directed edge contributes 1 to the out-degree of its tail and 1 to the in-degree of its head. Summing all out-degrees counts every edge once from the tail side, giving the edge count. Summing all in-degrees counts every edge once from the head side, giving the same count. Both totals equal the number of edges.

True, because the sum of degrees in any undirected graph equals 2|E|, which is always an even number since twice any integer is even.

False, because having all vertices of even degree does not guarantee connectivity—disconnected structures like several separate cycles can each have all-even degrees yet still form a disconnected graph.

True, because a vertex with out-degree 0 emits no edges and only receives incoming edges, which is exactly the definition of a sink in directed graph terminology.

In digraphs, each vertex v has two measures:

in-degree deg⁻(v): edges entering v

out-degree deg⁺(v): edges leaving v

Total degree = deg⁻(v) + deg⁺(v).

True, because a balanced digraph is defined by the condition deg⁺(v) = deg⁻(v) for every vertex v, representing systems with no accumulation or deficit of flow at any node.

True, because a loop in a directed or undirected setting simultaneously leaves and enters the same vertex, contributing +1 to in-degree and +1 to out-degree, giving a total degree contribution of 2.

Relation A must always hold because each directed edge increases exactly one vertex’s out-degree and exactly one vertex’s in-degree, meaning Σ deg⁺(v) = Σ deg⁻(v) = |E|, and none of the other listed relations consistently hold for all digraphs.

The weighted degree of v is 20 because in an undirected weighted graph the weighted degree is defined as the sum of the weights of all edges incident to v, so adding 4, 6, and 10 gives 20, which represents the total cost or capacity associated with that vertex.

Statements B and C are true because summing the degrees 2 + 2 + 3 + 3 + 4 + 4 yields 18, and dividing by 2 gives 9 edges according to the Handshaking Lemma, while statement A is false since a regular graph requires all degrees to be equal and statement D is false because there are actually four even-degree vertices (2, 2, 4, 4), not three.

The correct implication is that the junction accumulates 5 cars per hour because the total in-flow of 100 cars/hour exceeds the out-flow of 95 cars/hour, meaning a net +5 cars/hour build up at that vertex, which reflects the flow-balance principle used in network modeling where excess in-flow indicates accumulation.

A regular digraph with n vertices and constant out-degree r has n × r edges because each of the n vertices emits exactly r outgoing edges, and since each edge is counted exactly once at its tail, multiplying n by r gives the total number of directed edges in the graph.

The correct answer is 20 because in a digraph every directed edge contributes exactly one unit to some vertex’s in-degree, so the sum of all in-degrees equals |E| = 20, and by symmetry the total out-degree sum is also 20.

The correct statement is B because a directed edge A → B leaves A and therefore contributes +1 to A’s out-degree while entering B and contributing +1 to B’s in-degree, illustrating the rule that each directed edge increases one vertex’s out-degree and another vertex’s in-degree, with total in-degrees and total out-degrees both equaling the number of edges.